什么是计算笛卡尔积的好的非递归算法?
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03-07-2019 - |
题
注意
这不是特定于REBOL的问题。你可以用任何语言回答它。
背景
REBOL 语言支持创建称为“方言”的特定于域的语言。在REBOL 用语中。我已经为列表推导创建了这样的方言,这在REBOL中本身不受支持。
列表推导需要一个好的笛卡尔积算法。
问题
我已经使用元编程来解决这个问题,通过动态创建然后执行一系列嵌套的 foreach
语句。它工作得很漂亮。但是,因为它是动态的,所以代码不是很易读。 REBOL不能很好地进行递归。它迅速耗尽堆栈空间和崩溃。因此,递归解决方案是不可能的。
总而言之,我想用可读,非递归,“内联”替换我的元编程。算法,如果可能的话。解决方案可以使用任何语言,只要我可以在REBOL中重现它。 (我可以阅读任何编程语言:C#,C,C ++,Perl,Oz,Haskell,Erlang,等等。)
我应该强调,这个算法需要支持任意数量的集合才能“加入”,因为列表理解可能涉及任意数量的集合。
解决方案
这样的事情怎么样:
#!/usr/bin/perl
use strict;
use warnings;
my @list1 = qw(1 2);
my @list2 = qw(3 4);
my @list3 = qw(5 6);
# Calculate the Cartesian Product
my @cp = cart_prod(\@list1, \@list2, \@list3);
# Print the result
foreach my $elem (@cp) {
print join(' ', @$elem), "\n";
}
sub cart_prod {
my @sets = @_;
my @result;
my $result_elems = 1;
# Calculate the number of elements needed in the result
map { $result_elems *= scalar @这样的事情怎么样:
1 3 5
1 3 6
1 4 5
1 4 6
2 3 5
2 3 6
2 4 5
2 4 6
产生以下输出:
<*> } @sets;
return undef if $result_elems == 0;
# Go through each set and add the appropriate element
# to each element of the result
my $scale_factor = $result_elems;
foreach my $set (@sets)
{
my $set_elems = scalar @$set; # Elements in this set
$scale_factor /= $set_elems;
foreach my $i (0 .. $result_elems - 1) {
# Calculate the set element to place in this position
# of the result set.
my $pos = $i / $scale_factor % $set_elems;
push @{$result[$i]}, $set[ $pos ];
}
}
return @result;
}
产生以下输出:
<*>其他提示
使用速度提高3倍,使用的内存减少(回收次数减少)。
cartesian: func [
d [block! ]
/local len set i res
][
d: copy d
len: 1
res: make block! foreach d d [len: len * length? d]
len: length? d
until [
set: clear []
loop i: len [insert set d/:i/1 i: i - 1]
res: change/only res copy set
loop i: len [
unless tail? d/:i: next d/:i [break]
if i = 1 [break]
d/:i: head d/:i
i: i - 1
]
tail? d/1
]
head res
]
为了完整起见,这里罗伯特·甘博的答案被翻译成REBOL:
REBOL [] cartesian: func [ {Given a block of sets, returns the Cartesian product of said sets.} sets [block!] {A block containing one or more series! values} /local elems result row ][ result: copy [] elems: 1 foreach set sets [ elems: elems * (length? set) ] for n 0 (elems - 1) 1 [ row: copy [] skip: elems foreach set sets [ skip: skip / length? set index: (mod to-integer (n / skip) length? set) + 1 ; REBOL is 1-based, not 0-based append row set/(index) ] append/only result row ] result ] foreach set cartesian [[1 2] [3 4] [5 6]] [ print set ] ; This returns the same thing Robert Gamble's solution did: 1 3 5 1 3 6 1 4 5 1 4 6 2 3 5 2 3 6 2 4 5 2 4 6
这是一个Java代码,用于为任意数量的元素生成具有任意数量元素的笛卡尔积。
在此示例中列表“ls”包含4组(ls1,ls2,ls3和ls4),你可以看到“ls”和“ls”。可以包含任意数量的具有任意数量元素的集合。
import java.util.*;
public class CartesianProduct {
private List <List <String>> ls = new ArrayList <List <String>> ();
private List <String> ls1 = new ArrayList <String> ();
private List <String> ls2 = new ArrayList <String> ();
private List <String> ls3 = new ArrayList <String> ();
private List <String> ls4 = new ArrayList <String> ();
public List <String> generateCartesianProduct () {
List <String> set1 = null;
List <String> set2 = null;
ls1.add ("a");
ls1.add ("b");
ls1.add ("c");
ls2.add ("a2");
ls2.add ("b2");
ls2.add ("c2");
ls3.add ("a3");
ls3.add ("b3");
ls3.add ("c3");
ls3.add ("d3");
ls4.add ("a4");
ls4.add ("b4");
ls.add (ls1);
ls.add (ls2);
ls.add (ls3);
ls.add (ls4);
boolean subsetAvailabe = true;
int setCount = 0;
try{
set1 = augmentSet (ls.get (setCount++), ls.get (setCount));
} catch (IndexOutOfBoundsException ex) {
if (set1 == null) {
set1 = ls.get(0);
}
return set1;
}
do {
try {
setCount++;
set1 = augmentSet(set1,ls.get(setCount));
} catch (IndexOutOfBoundsException ex) {
subsetAvailabe = false;
}
} while (subsetAvailabe);
return set1;
}
public List <String> augmentSet (List <String> set1, List <String> set2) {
List <String> augmentedSet = new ArrayList <String> (set1.size () * set2.size ());
for (String elem1 : set1) {
for(String elem2 : set2) {
augmentedSet.add (elem1 + "," + elem2);
}
}
set1 = null; set2 = null;
return augmentedSet;
}
public static void main (String [] arg) {
CartesianProduct cp = new CartesianProduct ();
List<String> cartesionProduct = cp.generateCartesianProduct ();
for (String val : cartesionProduct) {
System.out.println (val);
}
}
}
use strict;
print "@1 3 5
1 3 6
1 4 5
1 4 6
2 3 5
2 3 6
2 4 5
2 4 6
输出
<*>\n" for getCartesian(
[qw(1 2)],
[qw(3 4)],
[qw(5 6)],
);
sub getCartesian {
#
my @input = @_;
my @ret = map [<*>
输出
<*>], @{ shift @input };
for my $a2 (@input) {
@ret = map {
my $v = <*>
输出
<*>;
map [@$v, <*>
输出
<*>], @$a2;
}
@ret;
}
return @ret;
}
输出
<*>编辑:此解决方案不起作用。 Robert Gamble是正确的解决方案。
我集思广益,想出了这个解决方案:
(我知道大多数人都不会知道REBOL,但它是一种相当可读的语言。)
REBOL [] sets: [[1 2 3] [4 5] [6]] ; Here's a set of sets elems: 1 result: copy [] foreach set sets [elems: elems * (length? set)] for n 1 elems 1 [ row: copy [] foreach set sets [ index: 1 + (mod (n - 1) length? set) append row set/(index) ] append/only result row ] foreach row result [ print result ]
此代码生成:
1 4 6
2 5 6
3 4 6
1 5 6
2 4 6
3 5 6
(首先阅读上面的数字,你可能认为有重复数据。我做过。但是没有。)
有趣的是,这段代码几乎使用了同样的算法(1 +((n - 1)%9)来破坏我的数字根问题。